Ta có: \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)
PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
______0,05_____0,1______0,05 (mol)
\(\Rightarrow m_{HCl}=0,1.36,5=3,65\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{3,65}{7,3\%}=50\left(g\right)\)
Ta có: m dd sau pư = 4 + 50 = 54 (g)
\(\Rightarrow C\%_{CuCl_2}=\dfrac{0,05.135}{54}.100\%=12,5\%\)
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\(n_{CuO}=\dfrac{4}{80}=0,05mol\\ CuO+2HCl\rightarrow CuCl_2+H_2O\)
0,05 0,1 0,05
\(m_{ddHCl}=\dfrac{0,1.36,5}{7,3}\cdot100=50g\\ C_{\%CuCl_2}=\dfrac{0,05.135}{50+4}\cdot100=12,5\%\)
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